Mechanics pleasee help acceleration

Bump

There's not a lot of point in bump'ing after a few minutes.

The ball was dropped and now it is slowing down, so direction of acceleration is upwards. (If it was downwards it would be getting faster.)

Accelerating upwards at 2.5 m/s/s

So, negative acceleration downwards -2.5 m/s/s

Thanks! And have you chosen your positive direction to be downwards/in the same direction as the direction of motion here?

Yes.

Yes.

Nothing too hard here
(i) Is simply getting the area under the curve you can do this with 3 triangles and a rectangle or 2 triangles and a trapezium.
(ii)Is about getting the gradient from 2 to 6 y2-y1/x2-x1 Specifying its negative as its decelerating.
(iii) With gradient from part (ii) you can use equation y-y1 = m(x-x1) to get the equation of the line (y is your velocty x is your time)

(IV) No air resistance as it has constant velocity

(V) Is asking for you to differentiate the equation and plug in the values of t=2 and t=6 and t=7 into the equation to check the gradient

(Vi) To calculate the distance you need to Integrate the equation for velocity and make your top band 7 and lower band 2.

Thats all good luck.

I thik you need to do more questions from the textbook before trying exam questions.

Depends on what sing you get:
+ will be upwards motion
- will be downwards motion

traditionally anyway. In practice you assume + to be anything you need, and - to be the opposite of that.

I am only stuck on the last question (part vi). And because the mark scheme is not clear enough just wanted to check if my answer of v=20-2.5t for part iii is correct.

Could you please kindly help or anyone else?

Thanks! I already get that though Could you please kindly lemme know if I've got the right ans for iii)? And for part vi I've explained below what I am stuck on. Could you please help? Thanks!!
So for iii keeping the positive direction downwards u=`20 a=-2.5 v=u +at
v=20-2.5t Is this correct?

for iv
v= -3/2 t^2 + 19/2 t +7 represents the velocity for the range of t values between t=2 and t=7, including t=2 and t=7
So at t=2 would the distance fallen be zero after using the integrated v equation - I am confused about this

iii) is correct
vi) requires integration I think. Integrating velocity gives displacement, and you use the boundary conditions to calculate the constant. Although when I look at the marking scheme they don't seem bothered about the constant.

Yes, the question asks you to use t=2 as the 'baseline' (so to speak), so this will also represent 0m of displacement.

Thank you so much I got the right answer. Just to confirm here is my working out:

Was wondering if you you could please kindly help me with doing it the mark scheme way as I don't undrrstand what they've done in the ms with all the random symbols

The funny symbols are as a result of your pdf reader of whatever other software you use to open the document not having the correct fonts installed (don't ask me how to fix it, I don't know lol). The software then replaces certain symbols with others from a different font, that unfortunately may look nothing like the original.

But if you use look closer... you can see that the "T" symbol appears to represent the "=" sign. "H" symbol looks like it means "-". "G" looks like it means "+". The stretched "O" appears to represent the integration symbol. etc.

I think the reason why it sometimes does and sometimes doesn't show the characters depends on whether or not there are spaces between the original characters. That's why sometimes it shows "=" and sometimes it shows "T".